3.614 \(\int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=555 \[ \frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{f \left (a^2+b^2\right )^2 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{f \left (a^2+b^2\right )^2 \sqrt {d \sec (e+f x)}}+\frac {2 (a \tan (e+f x)+b)}{f \left (a^2+b^2\right ) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {\left (2 a^2-3 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \left (a^2+b^2\right )^2 \sqrt {d \sec (e+f x)}}-\frac {5 a^2 b \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 f \left (a^2+b^2\right )^{5/2} \sqrt {d \sec (e+f x)}}+\frac {5 a^2 b \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 f \left (a^2+b^2\right )^{5/2} \sqrt {d \sec (e+f x)}}+\frac {5 a b^{3/2} \sqrt [4]{\sec ^2(e+f x)} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{9/4} \sqrt {d \sec (e+f x)}}-\frac {5 a b^{3/2} \sqrt [4]{\sec ^2(e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{9/4} \sqrt {d \sec (e+f x)}} \]

[Out]

5/2*a*b^(3/2)*arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^(9/4)/f/(d*s
ec(f*x+e))^(1/2)-5/2*a*b^(3/2)*arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(sec(f*x+e)^2)^(1/4)/(a^2
+b^2)^(9/4)/f/(d*sec(f*x+e))^(1/2)+(2*a^2-3*b^2)*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+
e)))*EllipticE(sin(1/2*arctan(tan(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(1/4)/(a^2+b^2)^2/f/(d*sec(f*x+e))^(1/2)-5/
2*a^2*b*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2),I)*(sec(f*x+e)^2)^(1/4)*(-tan(f*x+e)^2)^
(1/2)/(a^2+b^2)^(5/2)/f/(d*sec(f*x+e))^(1/2)+5/2*a^2*b*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b/(a^2+b^2)^
(1/2),I)*(sec(f*x+e)^2)^(1/4)*(-tan(f*x+e)^2)^(1/2)/(a^2+b^2)^(5/2)/f/(d*sec(f*x+e))^(1/2)-(2*a^2-3*b^2)*tan(f
*x+e)/(a^2+b^2)^2/f/(d*sec(f*x+e))^(1/2)+b*(2*a^2-3*b^2)*sec(f*x+e)^2/(a^2+b^2)^2/f/(d*sec(f*x+e))^(1/2)/(a+b*
tan(f*x+e))+2*(b+a*tan(f*x+e))/(a^2+b^2)/f/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.54, antiderivative size = 555, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3512, 741, 835, 844, 227, 196, 746, 399, 490, 1213, 537, 444, 63, 298, 205, 208} \[ \frac {5 a b^{3/2} \sqrt [4]{\sec ^2(e+f x)} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{9/4} \sqrt {d \sec (e+f x)}}+\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{f \left (a^2+b^2\right )^2 \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {5 a b^{3/2} \sqrt [4]{\sec ^2(e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{2 f \left (a^2+b^2\right )^{9/4} \sqrt {d \sec (e+f x)}}-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{f \left (a^2+b^2\right )^2 \sqrt {d \sec (e+f x)}}+\frac {2 (a \tan (e+f x)+b)}{f \left (a^2+b^2\right ) \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {\left (2 a^2-3 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \left (a^2+b^2\right )^2 \sqrt {d \sec (e+f x)}}-\frac {5 a^2 b \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 f \left (a^2+b^2\right )^{5/2} \sqrt {d \sec (e+f x)}}+\frac {5 a^2 b \sqrt {-\tan ^2(e+f x)} \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{2 f \left (a^2+b^2\right )^{5/2} \sqrt {d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2),x]

[Out]

(5*a*b^(3/2)*ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec[e + f*x]^2)^(1/4))/(2*(a^2 + b^2)
^(9/4)*f*Sqrt[d*Sec[e + f*x]]) - (5*a*b^(3/2)*ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(Sec
[e + f*x]^2)^(1/4))/(2*(a^2 + b^2)^(9/4)*f*Sqrt[d*Sec[e + f*x]]) + ((2*a^2 - 3*b^2)*EllipticE[ArcTan[Tan[e + f
*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/((a^2 + b^2)^2*f*Sqrt[d*Sec[e + f*x]]) - ((2*a^2 - 3*b^2)*Tan[e + f*x])/((a
^2 + b^2)^2*f*Sqrt[d*Sec[e + f*x]]) - (5*a^2*b*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), ArcSin[(Sec[e + f
*x]^2)^(1/4)], -1]*(Sec[e + f*x]^2)^(1/4)*Sqrt[-Tan[e + f*x]^2])/(2*(a^2 + b^2)^(5/2)*f*Sqrt[d*Sec[e + f*x]])
+ (5*a^2*b*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(Sec[e + f*x]^2)^(1/
4)*Sqrt[-Tan[e + f*x]^2])/(2*(a^2 + b^2)^(5/2)*f*Sqrt[d*Sec[e + f*x]]) + (b*(2*a^2 - 3*b^2)*Sec[e + f*x]^2)/((
a^2 + b^2)^2*f*Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])) + (2*(b + a*Tan[e + f*x]))/((a^2 + b^2)*f*Sqrt[d*Sec
[e + f*x]]*(a + b*Tan[e + f*x]))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 490

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]],
s = Denominator[Rt[-(a/b), 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), In
t[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 746

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(1/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(1/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1213

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

Rule 3512

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(d^(2
*IntPart[m/2])*(d*Sec[e + f*x])^(2*FracPart[m/2]))/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d \sec (e+f x)} (a+b \tan (e+f x))^2} \, dx &=\frac {\sqrt [4]{\sec ^2(e+f x)} \operatorname {Subst}\left (\int \frac {1}{(a+x)^2 \left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {\left (2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (-3+\frac {a^2}{b^2}\right )-\frac {a x}{2 b^2}}{(a+x)^2 \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)}}\\ &=\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {\left (2 b^3 \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {-\frac {a \left (a^2-4 b^2\right )}{2 b^4}-\frac {\left (2 a^2-3 b^2\right ) x}{4 b^4}}{(a+x) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}\\ &=\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {\left (5 a b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{(a+x) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (\left (2 a^2-3 b^2\right ) \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 b \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}\\ &=-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {\left (5 a b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (5 a^2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (\left (2 a^2-3 b^2\right ) \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{2 b \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}\\ &=\frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {\left (5 a b \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \sqrt [4]{1+\frac {x}{b^2}}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{4 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (5 a^2 \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (1+\frac {a^2}{b^2}-x^4\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}\\ &=\frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {\left (5 a b^3 \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {x^2}{a^2+b^2-b^2 x^4} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (5 a^2 b \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b^2}-b x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (5 a^2 b \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}\\ &=\frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}-\frac {\left (5 a b^2 \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}-b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (5 a b^2 \sqrt [4]{\sec ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}+b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}+\frac {\left (5 a^2 b \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a^2+b^2}-b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (5 a^2 b \cot (e+f x) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a^2+b^2}+b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{2 \left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}\\ &=\frac {5 a b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{2 \left (a^2+b^2\right )^{9/4} f \sqrt {d \sec (e+f x)}}-\frac {5 a b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) \sqrt [4]{\sec ^2(e+f x)}}{2 \left (a^2+b^2\right )^{9/4} f \sqrt {d \sec (e+f x)}}+\frac {\left (2 a^2-3 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {\left (2 a^2-3 b^2\right ) \tan (e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)}}-\frac {5 a^2 b \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^{5/2} f \sqrt {d \sec (e+f x)}}+\frac {5 a^2 b \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) \sqrt [4]{\sec ^2(e+f x)} \sqrt {-\tan ^2(e+f x)}}{2 \left (a^2+b^2\right )^{5/2} f \sqrt {d \sec (e+f x)}}+\frac {b \left (2 a^2-3 b^2\right ) \sec ^2(e+f x)}{\left (a^2+b^2\right )^2 f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}+\frac {2 (b+a \tan (e+f x))}{\left (a^2+b^2\right ) f \sqrt {d \sec (e+f x)} (a+b \tan (e+f x))}\\ \end {align*}

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Mathematica [C]  time = 33.44, size = 17812, normalized size = 32.09 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Sqrt[d*Sec[e + f*x]]*(a + b*Tan[e + f*x])^2),x]

[Out]

Result too large to show

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)^2), x)

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maple [B]  time = 3.72, size = 38644, normalized size = 69.63 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \sec \left (f x + e\right )} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate(1/(sqrt(d*sec(f*x + e))*(b*tan(f*x + e) + a)^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d/cos(e + f*x))^(1/2)*(a + b*tan(e + f*x))^2),x)

[Out]

int(1/((d/cos(e + f*x))^(1/2)*(a + b*tan(e + f*x))^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {d \sec {\left (e + f x \right )}} \left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))**(1/2)/(a+b*tan(f*x+e))**2,x)

[Out]

Integral(1/(sqrt(d*sec(e + f*x))*(a + b*tan(e + f*x))**2), x)

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